Problem Statement
At a depth of 7.5km in the ocean, the pressure is 75MPa. Assume a specific weight at the surface of 10 KN/m² and an average bulk modulus of elasticity of 2.5 GPa for that pressure range. Find:
- The change in specific volume between the surface and 7.5km
- The specific volume at 7.5km
- The specific weight at 7.5km
Given Data
- Pressure at 7.5km (P₂) = 75 MPa = 75 × 10⁶ N/m²
- Specific weight at surface (γ₁) = 10 KN/m² = 10,000 N/m²
- Bulk modulus (K) = 2.5 GPa = 2.5 × 10⁹ N/m²
- g = 9.81 m/s²
Solution
Part A: Change in Specific Volume
1. Calculate density at surface:
ρ₁ = γ₁/g = 10000/9.81 = 1019.4 kg/m³
2. Calculate specific volume at surface:
vs₁ = 1/ρ₁ = 1/1019.4 = 0.000981 m³/kg
3. Use bulk modulus formula for specific volume:
K = -P/(Δvs/vs₁)
Δvs = -(vs₁ × P)/K
Δvs = -(0.000981 × 75 × 10⁶)/(2.5 × 10⁹)
Δvs = -0.0000294 m³/kg
Change in specific volume = -0.0000294 m³/kg
Part B: Specific Volume at 7.5km
vs₂ = vs₁ + Δvs
vs₂ = 0.000981 – 0.0000294
vs₂ = 0.000951 m³/kg
Specific volume at 7.5km = 0.000951 m³/kg
Part C: Specific Weight at 7.5km
1. Calculate density at 7.5km:
ρ₂ = 1/vs₂ = 1/0.000951 = 1051.5 kg/m³
2. Calculate specific weight:
γ₂ = ρ₂ × g = 1051.5 × 9.81 = 10,315 N/m³
Specific weight at 7.5km = 10,315 N/m³
Key Points
• The negative change in specific volume indicates compression of the water with depth
• The increase in density and specific weight with depth reflects the compression of water under pressure
• The relatively small changes in these properties demonstrate water’s resistance to compression
